A vehicle starts moving along a straight line path from rest. In first

A vehicle starts moving along a straight line path from rest. In first `t` seconds it moves with an acceleration of 2 m/s² and then in next 10 seconds it moves with an acceleration of 5 m/s². The total distance travelled by the vehicle is 550 m. The value of time `t` is

10 s
13 s
20 s
25 s
This question was previously asked in
UPSC NDA-2 – 2024
The motion consists of two phases of constant acceleration. We need to find the time ‘t’ in the first phase such that the total distance covered in both phases is 550 m. By calculating the distance in each phase and setting the total distance equal to 550 m, we arrive at a quadratic equation in ‘t’. Solving this equation gives t = 10 s (the positive solution).
For the first phase (0 to t seconds): Initial velocity (u₁) = 0, acceleration (a₁) = 2 m/s². Distance s₁ = u₁t + (1/2)a₁t² = 0*t + (1/2)*2*t² = t². Velocity at time t (v₁) = u₁ + a₁t = 0 + 2t = 2t m/s.
For the second phase (t to t+10 seconds): Initial velocity (u₂) = v₁ = 2t m/s, acceleration (a₂) = 5 m/s², time (t₂) = 10 s. Distance s₂ = u₂t₂ + (1/2)a₂t₂² = (2t)*10 + (1/2)*5*(10)² = 20t + 250 m.
Total distance = s₁ + s₂ = t² + 20t + 250. Given total distance is 550 m, t² + 20t + 250 = 550, which simplifies to t² + 20t – 300 = 0.
Factoring the quadratic: (t + 30)(t – 10) = 0. Possible solutions are t = -30 or t = 10. Since time cannot be negative, t = 10 s.
We can verify the answer: If t=10 s, s₁ = 10² = 100 m. Velocity after 10s is v₁ = 2*10 = 20 m/s. In the next 10s, starting at 20 m/s with acceleration 5 m/s², distance s₂ = 20*10 + (1/2)*5*10² = 200 + 250 = 450 m. Total distance = s₁ + s₂ = 100 + 450 = 550 m, which matches the given information.
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